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英語 高校生

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名古屋大・文系 English words in length (Indicate the number of words you have written at the end of your answer. Do not count punctuation such as compras or periods as words 1 200 donors and delivers blood products to those who need them. Figure A below By year the Japanese Red Cross Society collects blood from voluntary shows how the mambers of younger (between the ages 16 and 39) and older between the ages 40 and 69) blood donors have changed in Japan from 2000 to 2019, as well as how the number of all blood donors has changed for the nineteen-year period. Figure B shows the total amount of blood donated in Linear trend lines are shown in dotted lines. Japan from 2000 to 2019 7.000.000 6.000.000 5.000.000- 4.000.000 3.000.000 2.000.000 1.000.000 Figure A Age (1639 years) A Age (40-69 years) .... ● All donors B. 1999 2001 2003 2005 2007 2000 2011 2013 2015 2017 2019 Years Amount of blood donated (liters) 2.000.000 2.000.000 1,500,000- 1.000.000- 500,000- Figure B QUESTIONS 2023 17 04 1999 2001 2003 2005 2007 2009 2011 2013 2015 2017 2019 Years Adapted from: Ministry of Health, Labour and Welfare website https://www.mhlw.go.jp/stf/seisakunitsuite/bunya/0000063233.html Write three 1. Describe what the Figure A show. trend lines in approximately 30 to 50 words. (Indicate the number of words you have written at the end of your answer Do not count punctuation such as commas or periods as words.) 2. Describe the trend depicted in Figure B. and explain how the amount of blood donated per donor has changed since 2000 by referring to both (Indicate the Figures A and B. Write approximately 30 to 50 words. Do not number of words you have written at the end of your answer. count punctuation such as commas or periods as words.)

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数学 高校生

ピンクのところどうしたらこのように展開できるんですか?

例題 344 内積と三角形の面積 点Oを原点とする.a=OA = (a1,a2), = OB = (b1,62), AOAB の面 積をSとする.このとき,次の式を示せ . せ s={√|ª³|b³²—(à• b)² = |a1b₁-a2b₁| BA A 考え方とのなす角を0とすると、△OAB の面積Sは, ■解答 S=OA-OB sine= |a|6|sine 5+36 9= 2 である. sin'0+cos0=1, d・L = |a||| cose を利用する aとのなす角を90°<9<180°) とすると, sin00 より, sin0=√1-cos' であるから, S=1/120A・OBsin=1/21|2|3|sine Focus -CO よって, ①, ②より, 与式は成り立つ. = |al|6|√1-cos²0=√|a³|b³(1-cos³0) - 100 = 1/2 √la 196³-|à P²|6|³ªcos²0 =√√ã³²|6³²-(¦â||b|cos0)² -√ã²b³²—(ã·¯)² また, lap=a²+a2²,16=622+62², at=ab+azb2 ①を成分で表す. であるから,①に代入して S=½ √(ai²+a2²)(b₁²+b₂²)— (a₁b₁+a2b2)² =1/12 -√(a₁b₂)²—2a₁b₁a₂b₂+(a₂b₁)² 1021 = 0 AO 8=58 ==√(a₁b₂-a₁b₁)² = |a₁b₁-a₂bil.... =3rd=d-0 0=A5+50+87 0=5+3+ HA 0 sin20+cos20=1 どのよ sin'0=1-cos20 sin0 >0 より sin0=√1-cos20 B △OAB で, OA= (a1,a2), OB=(b1,62) のとき, s=-=|a₁b₂-a₂b₁| lab2- 注 △ABCの面積も, a = AB, AC とおいて同様に求められる。 MASCH ATEA B O OH HA の結果を利用して、次の三角形の面積を求めよ. CADの面積 S b OS -MA) 38 (15-30-38-A ** a √A2=|A| S=absine 第9章

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